How do you show that a function is differentiable if there isn't a specific $x_{0}$ given?

Let's say we got a function given, simple example like $f(x)=x^{2}$. We are supposed to show this function is differentiable but there is no specific $x_{0}$ given where we check this.

What shall I do in this case?

Because I always used this formula to show it's differentiable:

$$\lim_{x\rightarrow x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}$$

and check if for left and right side there is same solution, if so, it's differentiable.

But I cannot use this formula because there is no $x_{0}$.

What then? Can I just derivate the function and if it works, conclude it is differentiable? Oh and additionally show that there is no definition-gap, jump,..?

Please do tell me what to do in such a case, I wouldn't know what to do. I smell a task like that in my exam.

• In questions like this one it is usually assumed you have to prove differentiability for any point $\;x_0\;$ in the domain of definition of the function, and in this case: in the whole real line. – DonAntonio Oct 13 '16 at 19:25

Just calculate the limit for any $\;x_0\;$ :

$$\lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}=\lim_{x\to x_0}\frac{(x-x_0)(x+x_0)}{x-x_0}=\lim_{x\to x_0}(x+x_0)=2x_0$$

• @cnmesr What "the number" are you talking about? I don't understand you... – DonAntonio Oct 13 '16 at 19:30
• Ohh really so easy, thanks! :) Never mind my comment, I understood it. – cnmesr Oct 13 '16 at 19:30
• Btw, it wouldn't have been enough if I showed that a function hasn't got any gaps, jumps and stuff like that to proof differentiability, right? – cnmesr Oct 13 '16 at 19:33
• @cnmesr Indeed: it isn't enough. Try with the function absolute value, $\;f(x)=|x|\;$ at $\;x_0=0\;$ . It hasn't jumps, gaps there but it isn't differentiable there. – DonAntonio Oct 13 '16 at 19:35
• Ok last question, sry now questions come to my mind... If I solve a task like that like in your answer, it is enough if I only do it once right? So I don't need to show it for left and right side? – cnmesr Oct 13 '16 at 19:37

You can't just take the derivative because you're assuming what you're trying to prove. That being said, here's what we do:

As mentioned in my comment, we'll use the $x-c$ definition (you actually wrote it incorrectly in your question).

We say that $f$ is differentiable at $c$ if the limit

$$\lim_{x \to c} \frac{f(x) - f(c)}{x-c}$$

exists. We call such a limit the derivative. Some books give this with $c$ relabeled as $x_0$ as you have done above.

So let $f(x) = x^2$. Then what? Well,

$$\frac{x^2 - c^2}{x-c} = x+c$$

and now you can take the limit as $x \to c$. Done!

• We didn't have this in our readings but I understand what you did now. Btw, what exactly did I write incorrectly in my question? Edit: Oh I see I have accidently written $\infty$ :o – cnmesr Oct 13 '16 at 19:31

Use the formula

$$f'(x) = \lim_{h \to 0} \left(\frac{f(x+h)-f(x)}{h}\right)$$

• Oh no, not this one.. I really don't like it :D There is no other way? ^.^ – cnmesr Oct 13 '16 at 19:25
• You can use the $x-c$ definition. I'll post an answer. – Sean Roberson Oct 13 '16 at 19:26